Question

Another limit problem! :D

Answer 1

try \(\large f(x) = e^{-sin(5x)}\)

Answer 2

a=0 i guess, hate to say but im not sure exactly wat the question is about...

Answer 3

hmm.. same here.. im not sure how to start too..

Answer 4

gotcha ! that function works !
and a = 0 fits as well..

Answer 5

uhm, how do i start or how should i approach this type of problem?

Answer 6

basically, the question is about finding the given limit.

by differentiating the f(x) at x = some value a

Answer 7

lets start wid simplifying the given limit

Answer 8

u shud knw below trig identity :-

cos(pi/2+x) = -sinx

Answer 9

yeah i do. so basically just differentiate the equation, and how did u know that a=0? or it can be any number?

Answer 10

\(\huge \lim_{ h->0} \frac{e^{\cos(\pi/2 + 5h)}-1}{h}\)

Answer 11

\(\huge\lim_{h->0} \frac{e^{-\sin(5h)}-1}{h}\)

Answer 12

\(\huge \lim_{h->0} \frac{e^{-\sin(5(0+h))}-e^0}{h}\)

Answer 13

fine so far ?

Answer 14

yep, got it (: then afterwards is just sub in h=0?

Answer 15

not exactly...
do we sense yet,

this limit is exactly the derivative of function \(\large f(x) = e^{-\sin5x}\) , at x = 0 ?

Answer 16

\(\large f(x) = e^{-\sin 5x}\)

find the derivative

\(\large f'(x) = e^{-\sin 5x}(-5\cos 5x)\)

evaluate at x = a= 0

\(\large f'(0) = e^{-\sin 0}(-5\cos 0)\)

\(\large f'(0) = -5\)

Answer 17

So,

\(\huge \lim_{h->0} \frac{e^{\cos(\pi/2+ 5h)}-1}{h} = -5\)

Answer 18

see if thats more/less confusing

Answer 19

hm, so the first part when you get f(x)= e^-sin5x, how did u exactly get this? by differentiating term e^cos(pi/2+5h)?

Answer 20

nopes,u knw the difference quotient right ?

Answer 21

\(\large \frac{f(x+\Delta x) - f(x)}{\Delta x}\)

Answer 22

above expression is called difference quotient,
u should be having lil experience in using it, to be able to sense that function backwards

Answer 23

ah that equation looks familiar. oh hm so u derive f(x) backwards? wow thats kinda tough haha

Answer 24

yeah its a tricky thing :)

Answer 25

i gtg.. cya !

Answer 26

thanks! (:

Answer 27

np :)